Category: Geometry

Question Number 199672 by Mingma last updated on 07/Nov/23 Answered by mr W last updated on 07/Nov/23 Commented by mr W last updated on 07/Nov/23…

Question Number 199510 by universe last updated on 04/Nov/23 Answered by mr W last updated on 06/Nov/23 Commented by mr W last updated on 06/Nov/23…

Question Number 199413 by universe last updated on 03/Nov/23 Answered by ajfour last updated on 03/Nov/23 $${b}\mathrm{cos}\:\alpha={R} \\ $$$$\left\{{R}\mathrm{tan}\:\alpha+\left(\frac{{a}}{{b}}\right){R}\right\}^{\mathrm{2}} +\left\{\left(\frac{{a}}{{b}}\right){R}\mathrm{tan}\:\alpha\right\}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${say}\:\:\:\mathrm{tan}\:\alpha={t} \\ $$$$\Rightarrow\:{t}^{\mathrm{2}}…

Question Number 199358 by ajfour last updated on 01/Nov/23 Commented by ajfour last updated on 01/Nov/23 $${Find}\:{R}. \\ $$ Answered by mr W last updated…

Question Number 199286 by ajfour last updated on 31/Oct/23 Answered by ajfour last updated on 31/Oct/23 $${Let}\:{height}\:{of}\://{gm}={h} \\ $$$${AB}=\mathrm{1}\:\: \\ $$$$\mathrm{2}{r}+\sqrt{\mathrm{1}−\mathrm{4}{r}^{\mathrm{2}} }={p} \\ $$$$\frac{{r}}{\mathrm{2}}\left(\mathrm{1}+{p}+\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }\right)=\frac{{p}}{\mathrm{2}}…